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Differential equation of solitary wave oscillons is defined by, $$ \Delta S -S +S^3=0 $$ How can we write this equation as, \begin{equation} \langle(\vec{\nabla}S)^2\rangle+\langle S^2\rangle-\langle S^4\rangle=0 \tag{1} \end{equation} where $\langle f\rangle:=\int d^Dx f(x)$. Furthermore, another virial identity can be found from the scaling transformation ($\vec{x}\to \mu \vec{x}$) by extremizing the scaled ($\vec{x}\to\mu \vec{x}$) of the action corresponding to $ \int d^Dx[(\vec{\nabla}S)^2+S^2-S^4/2]$ ( why $\frac{-S^4}{2}$): \begin{equation} (D-2)\langle(\vec{\nabla}S)^2\rangle+D\langle S^2\rangle-\frac{D}{2}\langle S^4\rangle=0 \tag{2} \end{equation} From Eqs. (1) and (2) one immediately finds \begin{equation} 2\langle S^2\rangle+\frac{1}{2}(D-4)\langle S^4\rangle=0\,, \end{equation} which equality can only be satisfied if $D<4$. D= Refers dimension.

I didn't understand the equation (2). Can someone do elaborately?

If you have any Query then ask me please. Thanks in advance.

I think this article will give you enough information: link check equations (21, 41, 42 43)

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$\def\mean#1{\left<#1\right>}$To obtain (1) multiply the differential equation by $S$ and integrate, noting that (if $S$ decays for $|x| \to \infty$) by partial integration $$ \int_{\mathbb R^D} \Delta S \cdot S \, dx = -\int_{\mathbb R^D} \nabla S\cdot \nabla S\, dx = \mean{(\nabla S)^2}$$ – martini May 22 '13 at 7:20
Dear martini, can you give me the source? And details calculations for the rest of order of (2)? – Complex Guy May 22 '13 at 7:25
This follows from the divergence theorem, and regarding (2) I do not quite understand where you want to scale? – martini May 22 '13 at 7:30
I think this article will give you enough information. check equations (21, 41, 42 43) – Complex Guy May 22 '13 at 7:32
If you have time then please look this question : Reference : equation no :(39,40) – Complex Guy May 22 '13 at 7:42
up vote 2 down vote accepted

The first part is simply multiplication by $S$, adding an integral sign, integration by parts for the first term.

For the second part consider a variation $\hat S(x;\mu)=S(x/\mu)$. Then do a change of variables $x=\mu y$. If the action is stationary under this spatial rescaling of $S$ then $d/d\mu$ at $\mu=1$ vanishes. Since the first term scales to $\mu^{D-2}$ with 2 from the derivative and $D$ from the Jacobian factor, this derivative gives a factor of $D-2$ as shown in the expression given. The rest are just Jacobian factors.

Edit: For example, $$ \left< (\nabla\hat S)^2\right> = \int d^Dx (\nabla_x S(x/\mu))^2 = \int \mu^D d^Dy (\mu^{-1} \nabla_y S(y))^2 = \mu^{D-2} \left< (\nabla S)^2\right>$$ The derivative at $\mu=1$ is exactly $(D-2) \left< (\nabla S)^2\right>$.

To clarify the first part, first multiply the equation by $-S$. Then integrate over all space. This gives the last two terms in the next line. Then for the first one, notice that $-S\nabla\cdot\nabla S$ can be integrated by parts using the divergence theorem. This shifts the $\nabla$ to the other term and changes the sign giving the shown result. If you prefer $\nabla\cdot(S\nabla S)$ can be expanded using the product rule and then the result substituted into the integral.

The factor of a half in the action arises from wanting the Euler-Lagrange equation for this action to be the given equation. The EL eqn contains $\partial L/\partial S$ so $L=S^2 + \alpha S^4$ results in a term $2S+ 4\alpha S^3$ so to get this agreeing with the eqn, up to an overall factor, choose $\alpha=-1/2$.

The deduction of (2) follows as I discuss above; rescale by $\mu$, then differentiate at $\mu=1$.

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Can you elaborate more please? – Complex Guy May 22 '13 at 10:03
Done, fixed typo too – Sharkos May 22 '13 at 10:20
So without $\mu=1$ we can not get the equation? what about other two? and how will I get $\begin{equation} 2\langle S^2\rangle+\frac{1}{2}(D-4)\langle S^4\rangle=0\,, \end{equation}$ after comparsion the equation (1) and(2) – Complex Guy May 22 '13 at 10:42
No idea what you mean. You need the action to be invariant under small changes when $S$ obeys the eq motion. For localized excitations, the small change achieved by scaling by $\mu$ near 1 results in a change in $S$ to the version with $\mu^{D-2} \cdot+\mu^D\cdot +\mu^D\cdot$. The change for small $\mu$ is the derivative at $\mu=1$, the point where the configuration is the one we started with. Then $(2)-(D-2)\times (1)$ gives you the answer; it's clearly just eliminating the gradient term. Are you sure you're ready to be doing this material? – Sharkos May 22 '13 at 11:50
Thank you I got it. – Complex Guy May 22 '13 at 13:18

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